使用 MySQL DB 的动态多级下拉列表 [英] Dynamic multi-level drop down using MySQL DB
问题描述
我正在尝试使用 PHP 创建一个多级下拉菜单，因为我之前没有 javascript 经验，但我也愿意学习.为了让每个人都容易理解，我以汽车为例.我希望用户选择一个品牌"示例福特，然后第二个下拉列表将填充福特制造的模型"，最后第三个下拉列表将填充福特汽车的颜色".我正在使用一个 MySQL 数据库，我想从它动态中提取所有数据，而不是对其中的值进行硬编码.我可以在第一个下拉列表中填充品牌"，但是当我选择品牌"时，第二个下拉不会填充新结果.我是 PHP 和 MySQL 的新手，但学习速度非常快.这是我的代码:
I am trying to create a multi level drop down menu hopefully using PHP since I have no prior experience in javascript but I am willing to learn that also. To make this easy for everyone to understand I am using cars as and example. I want the user to select a "Brand" example Ford, then a second drop down will populate with the "Models" Ford has made and finally a third drop down will populate with "Colors" of Ford vehicles. I am using a MySQL database that I want to pull all of the data from dynamically instead of hardcoding the values in. I am able to get the first drop down to populate with the "Brands" but when I select a "Brand" the second drop down does not populate with the new results. I am new to PHP and MySQL but a very quick learner. Here is my code:
INSERT_DROPDOWN.PHP
INSERT_DROPDOWN.PHP
<?php
$con = mysql_connect("localhost","username","XXXXXXXXXX");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("database", $con);
// Write out BRANDS.
$dquery = "SELECT brand FROM manufacture ORDER BY brand";
// Execute it, or return the error message if there's a problem.
$dresult = mysql_query($dquery) or die(mysql_error());
// Write out MODELS.
$dquery1 = "SELECT 'brand', 'model' FROM models WHERE brand='$dresult' ORDER BY model";
// Execute it, or return the error message if there's a problem.
$dresult1 = mysql_query($dquery1) or die(mysql_error());
// Write out COLORS.
$dquery2 = "SELECT color FROM color ORDER BY color";
// Execute it, or return the error message if there's a problem.
$dresult2 = mysql_query($dquery2) or die(mysql_error());
// if successful insert data into database, displays message "Successful".
if($dresult){
echo "Successful";
echo "<BR />";
}
else {
echo "ERROR1";
}
// close connection
mysql_close();
?>
测试.PHP
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<LINK href="CLL.css" rel="stylesheet" type="text/css">
<title>
testing
</title>
</head>
<body>
<?php
require "insert_dropdown.php";
?>
<p>
<?php
//Brand
$dropdown = "<select name='brand'>";
while($row = mysql_fetch_assoc($dresult))
{
$dropdown .= "\r\n<option value='{$row['brand']}'>{$row['brand']}</option>";
}
$dropdown .= "\r\n</select>";
echo $dropdown;
//Model
$dropdown1 = "<select name='brand'>";
while($row1 = mysql_fetch_assoc($dresult1))
{
$dropdown1 .= "\r\n<option value='{$row1['model']}'>{$row['brand']}</option>";
}
$dropdown1 .= "\r\n</select>";
echo $dropdown1;
//Color
$dropdown = "<select name='name'>";
while($row = mysql_fetch_assoc($dresult2))
{
$dropdown .= "\r\n<option value='{$row['color']}'>{$row['color']}</option>";
}
$dropdown .= "\r\n</select>";
echo $dropdown;
?>
</p>
</body>
</html>
推荐答案
我使用 AJAX 修复了代码
I fixed the code by using AJAX
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